ESAT Mathematics 2 Mock 3

Let $\displaystyle u = \sin\theta$ . The equation can be written in terms of $\displaystyle u$ as:For the expression to be defined, the denominator cannot be zero, so we must have $\displaystyle u+1 eq 0$ , which means $\displaystyle u eq -1$ . This implies that $\displaystyle \sin\theta eq -1$ .

The numerator can be factorised as a quadratic:Substituting this back into the equation gives:Since we have the condition $\displaystyle u eq -1$ , we can cancel the $\displaystyle (u+1)$ factor from the numerator and denominator:This gives $\displaystyle u = \dfrac{1}{2}$ . Substituting back $\displaystyle \sin\theta = u$ , we need to solve:In the interval $\displaystyle 0 \le \theta < 2\pi$ , the solutions are in the first and second quadrants.
The principal value is $\displaystyle \theta_1 = \dfrac{\pi}{6}$ .
The second solution is $\displaystyle \theta_2 = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ .

The sum of these solutions is:Therefore, the correct answer is B.

1. **Find the slope of $\displaystyle OA$ :**
The segment $\displaystyle OA$ connects $\displaystyle (0,0)$ to $\displaystyle (2a, a)$ , so its gradient is: $\displaystyle m = \dfrac{a - 0}{2a - 0} = \dfrac{1}{2}$ 2. **Find the coordinates of $\displaystyle C$ :**
Since side $\displaystyle CB$ is parallel to $\displaystyle OA$ , the line through $\displaystyle B(a, 3a)$ and $\displaystyle C$ has gradient $\displaystyle \dfrac{1}{2}$ : $\displaystyle y - 3a = \dfrac{1}{2}(x - a) \implies y = \dfrac{1}{2}x + \dfrac{5}{2}a$ Since $\displaystyle C$ lies on the $\displaystyle y$ -axis ( $\displaystyle x = 0$ ), $\displaystyle C = \left(0, \dfrac{5}{2}a\right)$ .

3. **Express the area of trapezium $\displaystyle OABC$ in terms of $\displaystyle a$ :**
Splitting the quadrilateral along the diagonal $\displaystyle OB$ into $\displaystyle \triangle OBC$ and $\displaystyle \triangle OAB$ :
- $\displaystyle \text{Area}(\triangle OBC) = \dfrac{1}{2} \times \text{base } OC \times \text{height } x_B = \dfrac{1}{2} \left(\dfrac{5}{2}a\right)(a) = \dfrac{5}{4}a^2$ - $\displaystyle \text{Area}(\triangle OAB) = \dfrac{1}{2} |(2a)(3a) - (a)(a)| = \dfrac{5}{2}a^2$ - Total area $\displaystyle = \dfrac{5}{4}a^2 + \dfrac{5}{2}a^2 = \dfrac{15}{4}a^2$ 4. **Solve for $\displaystyle a$ and find the coordinates of $\displaystyle B$ :** $\displaystyle \dfrac{15}{4}a^2 = 60 \implies a^2 = 16 \implies a = 4 \quad (\text{as } a > 0)$ Hence, the coordinates of $\displaystyle B(a, 3a)$ are $\displaystyle (4, 12)$ .

Let the arithmetic progression be denoted by $\displaystyle u_n$ , with first term $\displaystyle a=100$ and common difference $\displaystyle d=-6$ .

The $\displaystyle n$ -th term is given by the formula $\displaystyle u_n = a + (n-1)d$ .The sum of the first $\displaystyle n$ terms, $\displaystyle S_n$ , will increase as long as the terms being added are positive. The maximum sum will be achieved by summing all the positive terms. We need to find the largest integer $\displaystyle n$ for which $\displaystyle u_n > 0$ .The largest integer value for $\displaystyle n$ that satisfies this inequality is $\displaystyle n=17$ . Therefore, the maximum sum is the sum of the first 17 terms, $\displaystyle S_{17}$ .

We can calculate this sum using the formula $\displaystyle S_n = \dfrac{n}{2}(2a + (n-1)d)$ .To calculate $\displaystyle 17 \times 52$ : $\displaystyle 17 \times 50 = 850$ and $\displaystyle 17 \times 2 = 34$ . So, $\displaystyle 850 + 34 = 884$ .

Alternatively, we can find the last positive term, $\displaystyle u_{17} = 100 + 16(-6) = 100 - 96 = 4$ . Then use the formula $\displaystyle S_n = \dfrac{n}{2}(a+l)$ , where $\displaystyle l$ is the last term.Therefore the correct answer is 884.

The problem requires a two-step process: first find the x-coordinate of the point $\displaystyle P$ , and then find the equation of the curve to determine the corresponding y-coordinate.

Step 1: Find the x-coordinate of P

The gradient of the curve is given by $\displaystyle \dfrac{dy}{dx} = 3x^2$ . At the point $\displaystyle P$ , the gradient is $\displaystyle m$ . Let the coordinates of $\displaystyle P$ be $\displaystyle (x_P, y_P)$ .

We set the gradient function equal to $\displaystyle m$ :Solving for $\displaystyle x_P^2$ :Since we are given that $\displaystyle x > 0$ , we take the positive square root:Step 2: Find the equation of the curve

To find the equation of the curve, we integrate the gradient function with respect to $\displaystyle x$ :We are given that the curve passes through the origin $\displaystyle (0,0)$ . We use this information to find the constant of integration, $\displaystyle C$ :So, the equation of the curve is $\displaystyle y = x^3$ .

Step 3: Find the y-coordinate of P

Finally, we substitute the x-coordinate of $\displaystyle P$ into the equation of the curve:Using the law of indices $\displaystyle (a^b)^c = a^{bc}$ :Therefore the correct answer is A.

Because the cubic is monic and has stationary points at $\displaystyle x=1$ and $\displaystyle x=3$ ,Integrating givesso only the vertical-shift constant $\displaystyle C$ remains unknown. The stationary point at $\displaystyle x=1$ is a local maximum and the one at $\displaystyle x=3$ is a local minimum.

- A: $\displaystyle f(2)=C+2=5$ , so $\displaystyle C=3$ uniquely.
- B: $\displaystyle f(1)=C+4=1$ , so $\displaystyle C=-3$ uniquely.
- C: $\displaystyle f(2)=C+2=1$ , so $\displaystyle C=-1$ uniquely.
- D: $\displaystyle \int_0^1 f(x)\,dx=C+\dfrac{11}{4}=4$ , so $\displaystyle C=\dfrac54$ uniquely.
- F: the tangent at $\displaystyle x=0$ is $\displaystyle y=9x+C$ . Passing through $\displaystyle (1,5)$ gives $\displaystyle C=-4$ uniquely.

For E, the cubic has exactly two distinct real roots when either stationary point lies on the $\displaystyle x$ -axis. This occurs forgiving two possible values, $\displaystyle C=-4$ and $\displaystyle C=0$ . Therefore E is not sufficient, and the correct option is E.

Let the initial value of the function be $\displaystyle V = f(x)$ .

A decrease of 20% corresponds to multiplying the value by a factor of $\displaystyle 1 - 0.20 = 0.8$ .

When $\displaystyle x$ is increased by $\displaystyle d$ , the new value is $\displaystyle f(x+d) = V \times 0.8$ .

When $\displaystyle x$ is then increased by another $\displaystyle d$ (for a total increase of $\displaystyle 2d$ ), the value is multiplied by $\displaystyle 0.8$ again:Calculating the overall multiplier:This means the final value is 64% of the original value. The total percentage decrease is the difference from 100%:

We are given a linear relationship between $\displaystyle \ln y$ and $\displaystyle x$ . To find the connection to the constants $\displaystyle A$ and $\displaystyle b$ , we must linearise the equation $\displaystyle y = A b^{2x+3}$ by taking natural logarithms.Using the laws of logarithms, we can separate the terms:Rearranging this into the standard linear form $\displaystyle Y = mX + c$ , where $\displaystyle Y=\ln y$ and $\displaystyle X=x$ :We can now identify the gradient and the vertical intercept by comparing this equation to the given information.
The gradient is $\displaystyle m = 2\ln b$ . We are given $\displaystyle m=4$ , so $\displaystyle 2\ln b = 4$ , which means $\displaystyle \ln b = 2$ .
The vertical intercept is $\displaystyle c = \ln A + 3\ln b$ . We are given $\displaystyle c=7$ , so $\displaystyle \ln A + 3\ln b = 7$ .

Substituting $\displaystyle \ln b = 2$ into the intercept equation gives:This simplifies to $\displaystyle \ln A + 6 = 7$ , which gives $\displaystyle \ln A = 1$ . Therefore, $\displaystyle A = e^1 = e$ .

We notice the repeated term $\displaystyle \dfrac{x^2}{x+1}$ . Let $\displaystyle u = \dfrac{x^2}{x+1}$ . The equation becomes:Factorising gives $\displaystyle (u-5)(u+1) = 0$ , so $\displaystyle u=5$ or $\displaystyle u=-1$ . We must determine which of these values yield real roots for $\displaystyle x$ .

For $\displaystyle u=5$ , we have:The discriminant is $\displaystyle (-5)^2 - 4(1)(-5) = 45 > 0$ , confirming two distinct real roots. Their sum is $\displaystyle -(-5)/1 = 5$ .

For $\displaystyle u=-1$ , we have:The discriminant is $\displaystyle 1^2 - 4(1)(1) = -3 < 0$ , so there are no real roots here.

Thus, the sum of the real roots is simply $\displaystyle 5$ .

A circle tangent to both the $\displaystyle x$ -axis and the $\displaystyle y$ -axis that passes through the point $\displaystyle (1, 2)$ must be in the first quadrant. Its centre must therefore be at $\displaystyle (r, r)$ for some radius $\displaystyle r > 0$ .

The equation for such a circle is $\displaystyle (x-r)^2 + (y-r)^2 = r^2$ . Substituting the point $\displaystyle (1, 2)$ gives:Expanding and simplifying leads to a quadratic equation:This factorises to $\displaystyle (r-1)(r-5)=0$ , giving two possible values for the radius: $\displaystyle r=1$ and $\displaystyle r=5$ . The question asks for the largest radius, which is $\displaystyle 5$ .

We analyse each option to determine if it has real solutions.

For option E, let the equation be $\displaystyle 3^x + 3^{-x} = \dfrac{3}{2}$ .

Method 1: Substitution and discriminant.
Let $\displaystyle y = 3^x$ . Since $\displaystyle x$ is real, $\displaystyle y$ must be positive. The equation can be written in terms of $\displaystyle y$ :Multiplying by $\displaystyle 2y$ (which is non-zero) gives:This is a quadratic equation in $\displaystyle y$ . The discriminant is $\displaystyle \Delta = b^2 - 4ac = (-3)^2 - 4(2)(2) = 9 - 16 = -7$ .
Since the discriminant is negative, there are no real solutions for $\displaystyle y$ . Consequently, there are no real solutions for $\displaystyle x$ .

Method 2: Minimum value.
Let $\displaystyle f(x) = 3^x + 3^{-x}$ . Since $\displaystyle 3^x > 0$ for all real $\displaystyle x$ , we can use the AM-GM inequality:This implies $\displaystyle 3^x + 3^{-x} \ge 2$ . The minimum value of the left-hand side is 2, which occurs at $\displaystyle x=0$ . The right-hand side of the equation is $\displaystyle \dfrac{3}{2} = 1.5$ . Since $\displaystyle 1.5 < 2$ , the equation $\displaystyle 3^x + 3^{-x} = \dfrac{3}{2}$ can never be satisfied for any real $\displaystyle x$ .

Therefore, the equation in option E has no real solutions.

For the other options, a solution can be shown to exist, typically by considering the graphs of the functions on each side and using the Intermediate Value Theorem:
A: $\displaystyle f(x) = x^3+x-3$ . $\displaystyle f(1)=-1, f(2)=7$ . A root exists between 1 and 2.
B: $\displaystyle y=\ln(x)$ and $\displaystyle y=2-x$ must intersect. At $\displaystyle x=1$ , $\displaystyle \ln(1)=0 < 2-1=1$ . At $\displaystyle x=e$ , $\displaystyle \ln(e)=1 > 2-e \approx -0.718$ . A root exists between 1 and $\displaystyle e$ .
C: $\displaystyle y=e^x$ and $\displaystyle y=2-x^2$ must intersect. At $\displaystyle x=0$ , $\displaystyle e^0=1 < 2-0^2=2$ . At $\displaystyle x=1$ , $\displaystyle e^1 \approx 2.7 > 2-1^2=1$ . A root exists between 0 and 1.
D: $\displaystyle f(x) = \sin(x) - x + 1$ . $\displaystyle f(0)=1, f(2)=\sin(2)-1 < 0$ . A root exists between 0 and 2.

The derivative of the function is given as $\displaystyle f'(x) = 12x^2 - 12x$ .

To find the function $\displaystyle f(x)$ , we must integrate $\displaystyle f'(x)$ with respect to $\displaystyle x$ :where $\displaystyle C$ is an arbitrary constant of integration. The value of $\displaystyle C$ is not specified in the question.

Now, we evaluate the definite integral of $\displaystyle f(x)$ from $\displaystyle 0$ to $\displaystyle 1$ :We find the antiderivative of the expression in the integral:Now, we substitute the limits of integration:The value of the integral is $\displaystyle C - 1$ . Since the constant $\displaystyle C$ (which is equal to $\displaystyle f(0)$ ) is unknown, the value of the definite integral cannot be determined from the information given.

Therefore, the correct answer is F.

First, we find an expression for the general term $\displaystyle U_n$ of the second sequence.Using the logarithm law $\displaystyle \log(ab) = \log(a) + \log(b)$ , we get:We know that $\displaystyle \log_2(8) = 3$ and using the law $\displaystyle \log(x^y) = y\log(x)$ , we have:Since $\displaystyle \log_2(4) = 2$ , the expression for $\displaystyle U_n$ simplifies to:This is the formula for the $\displaystyle n$ -th term of an arithmetic sequence. We can find the first term $\displaystyle a$ and the common difference $\displaystyle d$ .

The first term (when $\displaystyle n=1$ ) is $\displaystyle a = U_1 = 2(1) + 1 = 3$ .
The second term (when $\displaystyle n=2$ ) is $\displaystyle U_2 = 2(2) + 1 = 5$ .
The common difference is $\displaystyle d = U_2 - U_1 = 5 - 3 = 2$ .

The sum of the first $\displaystyle N$ terms of an arithmetic sequence is given by the formula $\displaystyle S_N = \dfrac{N}{2}(2a + (N-1)d)$ .
We are given that this sum is 168. Substituting the values of $\displaystyle a$ and $\displaystyle d$ :This gives the quadratic equation:We can factorise this equation. We are looking for two numbers that multiply to $\displaystyle -168$ and add to $\displaystyle 2$ . These are $\displaystyle 14$ and $\displaystyle -12$ .This gives two possible solutions for $\displaystyle N$ : $\displaystyle N = -14$ or $\displaystyle N = 12$ .
Since the question asks for a positive integer $\displaystyle N$ , the correct value is $\displaystyle N=12$ .

Therefore the correct answer is 12.

A common mistake is to multiply both sides by $\displaystyle x-1$ . This is invalid because the sign of $\displaystyle x-1$ is unknown, so we don't know whether to reverse the inequality sign. Instead, we should rearrange the inequality to have zero on one side and combine terms into a single fraction.To make the numerator's leading term positive, we can multiply the fraction by $\displaystyle \dfrac{-1}{-1}$ or, equivalently, multiply the entire inequality by $\displaystyle -1$ and reverse the inequality sign.Factorising the numerator gives:The critical values, where the expression can change sign, are the roots of the numerator ( $\displaystyle x=-2, x=3$ ) and the root of the denominator ( $\displaystyle x=1$ ). These values divide the number line into four intervals.

We can test the sign of the expression in each interval:
- For $\displaystyle x > 3$ , all factors $\displaystyle (x-3)$ , $\displaystyle (x+2)$ , and $\displaystyle (x-1)$ are positive, so the expression is positive.
- For $\displaystyle 1 < x < 3$ , $\displaystyle (x-3)$ is negative, while $\displaystyle (x+2)$ and $\displaystyle (x-1)$ are positive. The expression is negative.
- For $\displaystyle -2 < x < 1$ , $\displaystyle (x-3)$ and $\displaystyle (x-1)$ are negative, while $\displaystyle (x+2)$ is positive. The expression $\displaystyle \dfrac{(-)(+)}{(-)}$ is positive.
- For $\displaystyle x < -2$ , all three factors are negative. The expression $\displaystyle \dfrac{(-)(-)}{(-)}$ is negative.

We are looking for where the expression is less than or equal to zero. This occurs for $\displaystyle x < -2$ and $\displaystyle 1 < x < 3$ . The inequality is non-strict ( $\displaystyle \le 0$ ), so we include the values where the numerator is zero ( $\displaystyle x=-2$ and $\displaystyle x=3$ ). We must exclude the value where the denominator is zero ( $\displaystyle x=1$ ).

Thus, the solution is $\displaystyle x \le -2$ or $\displaystyle 1 < x \le 3$ .

Bearings are measured clockwise from North. For the final position to be due North of the starting point, the resultant horizontal (East-West) displacement must be zero.

The horizontal component of a displacement $\displaystyle d$ on a bearing $\displaystyle \theta$ is $\displaystyle d \sin \theta$ . We sum the horizontal components of the two displacements:Substituting the exact values $\displaystyle \sin(60^\circ) = \dfrac{\sqrt{3}}{2}$ and $\displaystyle \sin(330^\circ) = -\dfrac{1}{2}$ :This simplifies to:Solving for $\displaystyle L$ gives $\displaystyle L = 12\sqrt{3}$ .

The expansion of $\displaystyle (x+3)^{10}$ is generated using the Binomial Theorem:The coefficient of the term $\displaystyle x^{10-k}$ is given by the integer product $\displaystyle \binom{10}{k} 3^k$ .
We require this coefficient to be divisible by 3.

We observe that for any $\displaystyle k \ge 1$ , the factor $\displaystyle 3^k$ is a multiple of 3. Thus, the coefficient is divisible by 3 for all $\displaystyle k \in \{1, 2, \dots, 10\}$ .

We must check the remaining case, $\displaystyle k=0$ , which corresponds to the leading term $\displaystyle x^{10}$ :This is not divisible by 3. Consequently, exactly 10 of the 11 coefficients in the expansion are divisible by 3.

First, calculate the area of the trapezium: $\displaystyle \text{Area}_{\text{trapezium}} = \dfrac{1}{2} \left( \dfrac{3x - 1}{4} + \dfrac{x + 5}{2} \right) \times 6 = 3 \left( \dfrac{3x - 1 + 2(x + 5)}{4} \right) = \dfrac{3(5x + 9)}{4} = \dfrac{15x + 27}{4}$ Next, calculate the area of the right-angled triangle: $\displaystyle \text{Area}_{\text{triangle}} = \dfrac{1}{2} \times \left( \dfrac{5x - 2}{3} \right) \times 9 = \dfrac{3(5x - 2)}{2} = \dfrac{15x - 6}{2} = \dfrac{30x - 12}{4}$ The area of the trapezium is strictly greater than the area of the triangle when: $\displaystyle \dfrac{15x + 27}{4} > \dfrac{30x - 12}{4}$ $\displaystyle 15x + 27 > 30x - 12$ $\displaystyle 39 > 15x \implies x < \dfrac{39}{15} = \dfrac{13}{5}$ Additionally, all side lengths must be strictly positive:
1. $\displaystyle \dfrac{3x - 1}{4} > 0 \implies x > \dfrac{1}{3}$ 2. $\displaystyle \dfrac{x + 5}{2} > 0 \implies x > -5$ 3. $\displaystyle \dfrac{5x - 2}{3} > 0 \implies x > \dfrac{2}{5}$ Since $\displaystyle \dfrac{2}{5} > \dfrac{1}{3}$ , the strictest lower bound is $\displaystyle x > \dfrac{2}{5}$ .

Combining the two conditions gives the complete set of values: $\displaystyle \dfrac{2}{5} < x < \dfrac{13}{5}$